Solve 2+y(1-3y)=y(y-3)

$$2 + y ( 1 - 3 y ) = y ( y - 3 )$$

Solve for y

$$y = \frac{\sqrt{3} + 1}{2} \approx 1.366025404$$

$$y=\frac{1-\sqrt{3}}{2}\approx -0.366025404$$

Steps Using the Quadratic Formula

Use the distributive property to multiply y by 1-3y.

$$2+y-3y^{2}=y\left(y-3\right)$$

Use the distributive property to multiply y by y-3.

$$2+y-3y^{2}=y^{2}-3y$$

Subtract y^{2} from both sides.

$$2+y-3y^{2}-y^{2}=-3y$$

Combine -3y^{2} and -y^{2} to get -4y^{2}.

$$2+y-4y^{2}=-3y$$

Add 3y to both sides.

$$2+y-4y^{2}+3y=0$$

Combine y and 3y to get 4y.

$$2+4y-4y^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-4y^{2}+4y+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 4 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-4±\sqrt{4^{2}-4\left(-4\right)\times 2}}{2\left(-4\right)}$$

Square 4.

$$y=\frac{-4±\sqrt{16-4\left(-4\right)\times 2}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$y=\frac{-4±\sqrt{16+16\times 2}}{2\left(-4\right)}$$

Multiply 16 times 2.

$$y=\frac{-4±\sqrt{16+32}}{2\left(-4\right)}$$

Add 16 to 32.

$$y=\frac{-4±\sqrt{48}}{2\left(-4\right)}$$

Take the square root of 48.

$$y=\frac{-4±4\sqrt{3}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$y=\frac{-4±4\sqrt{3}}{-8}$$

Now solve the equation y=\frac{-4±4\sqrt{3}}{-8} when ± is plus. Add -4 to 4\sqrt{3}.

$$y=\frac{4\sqrt{3}-4}{-8}$$

Divide -4+4\sqrt{3} by -8.

$$y=\frac{1-\sqrt{3}}{2}$$

Now solve the equation y=\frac{-4±4\sqrt{3}}{-8} when ± is minus. Subtract 4\sqrt{3} from -4.

$$y=\frac{-4\sqrt{3}-4}{-8}$$

Divide -4-4\sqrt{3} by -8.

$$y=\frac{\sqrt{3}+1}{2}$$

The equation is now solved.

$$y=\frac{1-\sqrt{3}}{2} y=\frac{\sqrt{3}+1}{2}$$

Steps for Completing the Square

Use the distributive property to multiply y by 1-3y.

$$2+y-3y^{2}=y\left(y-3\right)$$

Use the distributive property to multiply y by y-3.

$$2+y-3y^{2}=y^{2}-3y$$

Subtract y^{2} from both sides.

$$2+y-3y^{2}-y^{2}=-3y$$

Combine -3y^{2} and -y^{2} to get -4y^{2}.

$$2+y-4y^{2}=-3y$$

Add 3y to both sides.

$$2+y-4y^{2}+3y=0$$

Combine y and 3y to get 4y.

$$2+4y-4y^{2}=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$4y-4y^{2}=-2$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-4y^{2}+4y=-2$$

Divide both sides by -4.

$$\frac{-4y^{2}+4y}{-4}=-\frac{2}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$y^{2}+\frac{4}{-4}y=-\frac{2}{-4}$$

Divide 4 by -4.

$$y^{2}-y=-\frac{2}{-4}$$

Reduce the fraction \frac{-2}{-4} to lowest terms by extracting and canceling out 2.

$$y^{2}-y=\frac{1}{2}$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-y+\left(-\frac{1}{2}\right)^{2}=\frac{1}{2}+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-y+\frac{1}{4}=\frac{1}{2}+\frac{1}{4}$$

Add \frac{1}{2} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-y+\frac{1}{4}=\frac{3}{4}$$

Factor y^{2}-y+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{1}{2}\right)^{2}=\frac{3}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{1}{2}\right)^{2}}=\sqrt{\frac{3}{4}}$$

Simplify.

$$y-\frac{1}{2}=\frac{\sqrt{3}}{2} y-\frac{1}{2}=-\frac{\sqrt{3}}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$y=\frac{\sqrt{3}+1}{2} y=\frac{1-\sqrt{3}}{2}$$